Explanation:
The given equation is as follows.
[tex]PbBr_2 \rightleftharpoons Pb^{2+} + 2Br^{-}[/tex]
s 2s
It is given that,
[tex]K_{sp} = [Pb^{2+}][Br^{-}]^{2} = 6.60 \times 10^{-6}[/tex]
Let the solubility of given ions be "s".
Since, KBr on dissociation will given bromine ions.
Hence, [tex]K_{sp} = [Pb^{2+}] \times ([Br^{-}])^{2}[/tex]
[tex]6.60 \times 10^{-6}[/tex] = [tex]s \times (2s)^{2}[/tex]
= [tex]1.18 \times 10^{-2}[/tex] M
Therefore, solubility of [tex][PbBr_{2}][/tex] is [tex]1.18 \times 10^{-2}[/tex] M in KBr.
Now, we will calculate the molar solubility of [tex]PbBr_{2}[/tex] in 0.5 M KBr solution as follows.
[tex]K_{sp} = (s) \times (2s + 0.5)^{2}[/tex]
[tex]6.60 \times 10^{-6}[/tex] = [tex]4s^{3} + 0.25s + 2s^{2}[/tex]
s = [tex]2.63 \times 10^{-5}[/tex]
Thus, we can conclude that molar solubility of [tex][PbBr_{2}][/tex] in 0.500 m KBr solution is [tex]2.63 \times 10^{-5}[/tex].
